00:01
So our question says an educational researcher devised a wooden toy assembly project to test learning in six year old at the time in seconds to assembly day to assemble the project was noted and the toy was disassembled out of the children's sites then the children was given the child was given the tax to repeat the researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times so at the level of significance of 0 .0 can it be that in learning time that learning took place i was supposed to use the p -value method and a 99 percent confidence so we have our data set for seven children we have trial one we have trial two so for trial one uh i decide to call it a sample one s one and uh for trial two i decide to call it sample two s two so our n1 is equal to seven our n2 is equal to seven so we need a sample mean of the first trial the sample variance of the first trial the sample mean of the second trial and the sample variance of the second trial so uh we have our datasets for the first trial right here and the data set for the second trial over here so i was able to extract each of these trial data and then slot it into a calculator that will help us get the sample mean and the sample variance so this is the first trial data so i click on calculate sample mean is 125 .43 so this is our 125 .43 and the sample variance in this case of us we have 336 .29 so let's go for the second sample data so this is it that's the second trial we click on calculates we have our sample to be 113 .57 and the sample variance we have that to be equals to 472 .62 so the first step we are supposed to get the we are supposed to make our decisions using the hypothesis test and also using a 99 % confidence in time so our first step is to perform our hypothesis test the null and alternative hypothesis so h notice is on the fact that mu one is equal to m2 and the alternative is that me one is lesser than mule so that being said the next step is to perform our test statistics and as you can see we have a very little sample sizes our sample sizes are very very less we are going to be using a t test as our test statistics so we have our t test to be t is equals to x1 bar minus x2 bar divided by the square root of s1 squared divided by n1 plus s2 squared by n2 please note that we are also we this is a formula for independent sample t test because each samples are actually independent on each other so our x1 bar minus x2 bar we have a 125 .43 minus 113 .5 .57 divided by the square root of so we have 366 .29 divided by 7 plus 472 .62 divided by 7 so let's do the math we have a 125 .43 minus 113 .5 .57.
03:38
We have a 133 .7.
03:38
We have a that gives us 11 .86 divided by so we have a less use a calculator to work on the square roots of 366 .09 divided by 7 plus 472 .62 divided by 7 and we have this to be across to 10 .947 so when we do the math we have 11 .86 divided by 10 .947 and our test statistics is one point excuse me our test statistics is 1 .083 so the next step is to get the corresponding p value of our statistics and to do so we'll be needing the digger freedom because you are working on a t test so we need the degree freedom and for an independent sample to test the graph freedom is n1 plus n2 minus 2 so that's going to be 7 plus 7 minus 2 that's 14 minus 2 which gives us 12 our test statistics t is a 1 .083 so let's go to a calculator i'll be using a calculator to get the corresponding p value so this is a p value calculator our test is a left still test we have it's a score the gear freedom is 12 and we have our test statistics to be 1 .083 and that's 0 .89 9 .9 so our p value is 0 .849 so we have our p value our level of significance is 0 .0 now it's time to make our decision so for the decision rule i'll be using the p value method and the rule this decision rule states that if the p value is greater than alpha we fail to reject the null hypothesis and if the p value is lesser than alpha we actually reject reject the null hypothesis as you can see 0 .849 is greater than 0 .05 hence our decision is that we fail to reject the null hypothesis.
05:52
So that means in conclusion there is actually no enough evidence to support the claim that learning took place.
05:59
So this is also solving our question using the concept of hypothesis tests.
06:04
So let's solve the same question but just by constructing a confidence interval.
06:08
Then after constructing our interval we are going to be using the interval to solve the same question we are supposed to arrive at the same decision so to construct a 99 % confidence in 99 % confidence in favor for the difference between the population means of both sample data we have that mu one minus mu two is equals to x1 bar minus x2 bar plus or minus we have the critical value times the square root of we have a s1 squared divided by n2 so the only problem we have here is getting the critical value which can be getting by using a critical value calculator so we have a two students so a confidence interval is seen as a two -tail test the degree of freedom is going to be 12 and our level of significance is going to be 0 .01 and that gives us 3 .0545 so we have a 125 .43 minus 113 .5 .57 plus or minus 3 .055 4 5 times okay i guess i'm correct 3 .0545 exactly time so when we've done the curves of everything here before we had our answer to be equals to 10 .947 so we have this to be 11 .86 plus or minus so 10 .947 times 3 .0545 that gives us a 33 .4365 0 .4365 okay so we can split up the data set we can have 11 .86 plus 33 .365 or we can say 11 .86 minus 33 .365 so when we do the math here we have a 45 .2985 or one will subtract that we have a minus 21 .57.
08:14
It's seven, okay? so our 99 % confidence interval for the difference between the population mean is a minus 21 .5787.
08:25
47 .45 .45.
08:29
So we want to use this confidence interval to make our decision.
08:32
Now, if there is a, if zero is a member of our confidence interval, it simply means that there there's actually no difference between both samples...